Displacement Current: The Term That Predicted Light

Charge is conserved — locally

Start from a law you already trust: conservation of charge. But state it sharply. Charge does not merely stay constant overall — it cannot even disappear here and reappear there. If the charge in a region drops, an equal current must have flowed out through the boundary. That local bookkeeping is the continuity equation.

\dfrac{\partial \rho}{\partial t} + \nabla\cdot\mathbf{J} = 0

The continuity equation: the rate charge density falls at a point equals the current diverging away from it.

The crack: Ampère's law fails for a charging capacitor

Now watch the original Ampère law break. Take a wire charging a capacitor and a loop encircling the wire. Ampère says \oint\mathbf{B}\cdot d\boldsymbol{\ell}=\mu_0 I_{\text{enc}}, where I_{\text{enc}} is the current through any surface bounded by the loop. Cap the loop with a flat disk pierced by the wire: current I. Now cap it instead with a balloon-shaped surface that bulges between the capacitor plates, where no charge crosses: current $0$. Same loop, two answers.

Maxwell's fix

Maxwell's insight: between the plates there is no moving charge, but there is a rapidly changing electric field as the capacitor charges. He proposed that a changing \mathbf{E} acts as a source of \mathbf{B} just as a real current does. He called \varepsilon_0\,\partial\mathbf{E}/\partial t the displacement current. Through the bulging surface, its flux is exactly the I that was 'missing' — consistency restored.

\mathbf{J}_d = \varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}, \qquad I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}

The displacement current density and the displacement current through a surface — a changing E-flux, not a flow of charge.

The fix is not a lucky patch; it is forced. Take the divergence of the full Ampère–Maxwell law. The left side vanishes identically (the divergence of any curl is zero), and what survives on the right is precisely the continuity equation. Maxwell's term is exactly the one that makes the magnetic law compatible with charge conservation — no more, no less.

0 = \nabla\cdot(\nabla\times\mathbf{B}) = \mu_0\Big(\nabla\cdot\mathbf{J} + \varepsilon_0\dfrac{\partial(\nabla\cdot\mathbf{E})}{\partial t}\Big) = \mu_0\Big(\nabla\cdot\mathbf{J} + \dfrac{\partial\rho}{\partial t}\Big)

Taking the divergence of Ampère–Maxwell regenerates the continuity equation — charge conservation is now built in.

The payoff: light

With the new term in place, empty space is no longer inert. In a region with no charges or currents, take the curl of Faraday's law, use the identity \nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E}, and substitute Ampère–Maxwell for \nabla\times\mathbf{B}. Because \nabla\cdot\mathbf{E}=0 in vacuum, the first piece drops and a wave equation is left standing.

Worked derivation. (1) \nabla\times(\nabla\times\mathbf{E}) = -\dfrac{\partial}{\partial t}(\nabla\times\mathbf{B}). (2) The left side is \nabla(\nabla\cdot\mathbf{E})-\nabla^2\mathbf{E} = -\nabla^2\mathbf{E} in vacuum. (3) The right side is -\dfrac{\partial}{\partial t}\big(\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t\big). (4) Equate: \nabla^2\mathbf{E} = \mu_0\varepsilon_0\,\partial^2\mathbf{E}/\partial t^2. A perfectly symmetric calculation gives the same equation for \mathbf{B}.

\nabla^2\mathbf{E} - \mu_0\varepsilon_0\dfrac{\partial^2 \mathbf{E}}{\partial t^2} = 0, \qquad v = \dfrac{1}{\sqrt{\mu_0\varepsilon_0}} = c

The wave equation for E in vacuum, propagating at exactly the speed of light. The B-field obeys an identical equation.

Without that one extra term there is no wave — the fields would sit inert. With it, Maxwell predicted electromagnetic waves on paper decades before Hertz generated and detected radio waves in the lab. A misplaced sign or a missing \varepsilon_0 and the whole electromagnetic spectrum would evaporate. That is how much rides on displacement current.